1000 KVA 11kV/415V RTR = 1.%, XTR -6.0% 200 KVAR 16 -C VLL = 11 kV, 50 Hz MVASc = 200 X/R = 2.4 Linear and nonlinear plant loads Figure Q3(c)
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- The symbol shown is a(n) a. iron core transformer. b. auto transformer. c. current transformer. d. air core transformer.A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding short-circuited, rated current flows when the voltage applied to the primary is 5.5 kV. The power input is read as 100 kW. Determine Req1andXeq1 in ohms referred to the high-voltage winding.A single-phase l0-kVA,2300/230-volt,60-Hz two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 to 2530 volts (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of this kVA rating is transformed by magnetic induction? (C) The following data are obtained from tests carried out on the transformer when it is connected as a two-winding transformer: Open-circuit test with the low-voltage terminals excited: Applied voltage =230V, input current =0.45A, input power =70W. Short-circuit test with the high-voltage terminals excited: Applied voltage =120, input current =4.5A, input power =240W. Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransformer than as a two-winding transformer.
- In developing per-unit circuits of systems such as the one shown in Figure 3.10. when moving across a transformer, the voltage base is changed in proportion to the transformer voltage ratings. (a) True (b) FalseConsider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in , place the polarity marks, and label the terminals a, b, and c in accordance with the American standard. (b) Relabel the terminals a, b, and c such that VAN is 90 out of phase with Va for positive sequence.A single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R1=0.75 Resistance of the 240-V winding: R2=0.0075 Leakage reactance of the 2400-V winding: X1=1.0 Leakage reactance of the 240-V winding: X2=0.01 Exciting admittance on the 240-V side =0.003j0.02S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.
- In the single-phase, three- winding, transformer, the primary current is equal to lo at the condition of ints there is no load on tertiary windings the current in secondary windings equal to zero the resistances (R1, R2, and R3) in transformer windings are equal the inductances (L1, L2, and L3) in transformer windings are equal.Q3/A/ The equivalent circuit impedances of a 20-kVA, 8000/240 V, 60-Hz transformer are to be determined. The open-circuit test was performed on the secondary side of the transformer and the short circuit test was performed on the primary side of the transformer .The following data was taken: Open-circuit Short-circuit test test (on secondary) Voc 240 v (on primary) Vsc 489 v Isc 2.5 A Psc 240 W loc 7.133 A Poc 400 WThe ratings of a single-phase three-winding transformer are as follows:*primary winding: 300 MVA, 14kV*secondary winding: 300MVA, 200kV*tertiary winding: 40MVA, 20kV. The leakage reactances are:*Xps = 0.1 at 300MVA, 14kV*Xpt = 0.16 at 40MVA, 14kV*Xst = 0.14 at 50MVA, 200kV Neglecting the winding resistance and exciting current, calculate Xp, Xs and Xt using 300 MVA and 14kV as the base for the primary winding.
- What is the relationship between primary induced emf( E1) and applied voltage (V1) of a Transformer? O a. Both are unequal and in phase O b. Both are unequal and 180 degree out of phase O c. Both are equal and 180 degree out of phase O d. Both are equal and inphaseTwo transformers A and B have the following data Transformer A: 6600/440 V, (250 KVA ) requires ( 6%) of rated voltage to circulate (90%) of full load current at power factor (0.25) when the low voltage winding is short circuited and the no load loss is equal to (2500 W). Transformer B: 6600/420 V, (600 KVA) requires (9%) of rated voltage to circulate (80%) of full load current at power factor (0.16) when low voltage winding is short circuited and no load loss is equal to (4500 W). Calculate the followings: If both transformers are connected in parallel to supply a load (1.2 + j 2.3) ohm, find the output power of each transformer.Two transformers A and B have the following data Transformer A: 6600/440 V, (250 KVA ) requires ( 6%) of rated voltage to circulate (90%) of full load current at power factor (0.25) when the low voltage winding is short circuited and the no load loss is equal to (2500 W). Transformer B: 6600/420 V, (600 KVA) requires (9%) of rated voltage to circulate (80%) of full load current at power factor (0.16) when low voltage winding is short circuited and no load loss is equal to (4500 W). Calculate the followings: The terminal voltage when transformer A is supplying half load at power factor (0.6) lagging and transformer B is supplying (85%) of full load at power factor (0.85) leading.